Question

A survey of 280 homeless persons showed that 63 were veterans. Construct a 90% confidence interval for the proportion of homeless persons who are veterans.

a. (0.184, 0.266)
b. (0.176, 0.274)
c. (0.167, 0.283)
d. (0.161, 0.289)

Answer 1

To construct the 90% confidence interval for the proportion of homeless persons who are veterans, we used the formula for the population proportion confidence interval, calculated the sample proportion, found the z-value for a 90% confidence level (1.645), calculated the margin of error, and then determined the interval to be approximately (0.183, 0.267), with option (a) being the closest match.

Step-by-step explanation:

To construct a 90% confidence interval for the proportion of homeless persons who are veterans, we use the following formula for a confidence interval for a population proportion:
CI = p± Z*sqrt[p(1-p)/n]

Where:

• p is the sample proportion, which is the number of veterans divided by the total number of homeless persons surveyed.
• Z* is the z-value that corresponds to the desired level of confidence.
• n is the sample size.

For this question:

• p = 63/280
• n = 280

To find the z-value for a 90% confidence level, we refer to a z-table or use statistical software. The z-value for 90% confidence is approximately 1.645.

Let's calculate the confidence interval:

p = 63/280 = 0.225

Z* = 1.645

Standard Error (SE) = sqrt[p(1-p)/n] = sqrt[0.225(1-0.225)/280] = 0.0253

Multiplying Z* by SE gives us the margin of error (MOE):

MOE = Z* * SE = 1.645 * 0.0253 = 0.0416

Subtracting and adding MOE from p we get the confidence interval:

CI = (0.225 - 0.0416, 0.225 + 0.0416) = (0.1834, 0.2666)

After rounding to three decimal places:

CI = (0.183, 0.267)

Option (a), (0.184, 0.266), closest matches our calculated interval after rounding.

Answer 2

`0.225 - 1.64 \sqrt{(0.225(1-0.225))/(280)}=0.184`

`0.225 + 1.64 \sqrt{(0.225(1-0.225))/(280)}=0.266`

And the best option would be:

a. (0.184, 0.266)

Step-by-step explanation:

We have the following info given:

`X= 63` represent the homeless persons that were veterans

`n= 280` represent the sampel size

The estimated proportion for this case would be:

`\hat p=(63)/(280)= 0.225`

The confidence interval would be given by this formula

`\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}`

For the 90% confidence interval the value of
`\alpha=1-0.9=0.1` and
`\alpha/2=0.05`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.64`

And replacing into the confidence interval formula we got:

`0.225 - 1.64 \sqrt{(0.225(1-0.225))/(280)}=0.184`

`0.225 + 1.64 \sqrt{(0.225(1-0.225))/(280)}=0.266`

And the best option would be:

a. (0.184, 0.266)